229 is another term because 613^2 divides 229!+1. See A115091 for primes whose square divides m!+1 for some m. An examination of the factorizations of m!+1 for m<=100 found no additional squares. - T. D. Noe, Mar 01 2006
562 is also a term because 562!+1 is divisible by 563^2. - Vladeta Jovovic, Mar 30 2004
Comment from Francois BRUNAULT, Nov 23 2008: A web search reveals that for 1 <= k <= 228 there are 82 values of k for which k! + 1 has not been completely factored (the smallest is k=103), so showing that 229 and 562 are indeed the next two terms will be a huge task. I checked that k!+1 is not divisible by p^2 for k <= 1000 and prime p < 10^8.
It is very likely that 229 and 562 are the next two terms, but this has not yet been proved. - Nov 29 2008
Contains A007540(n)-1 for all n. That sequence is conjectured to be infinite. - Robert Israel, Jul 04 2016
This sequence includes A146968 (solutions of Brocard's problem). - Salvador Cerdá, Mar 08 2016
I may be wrong, but perhaps it is not yet proved that there are no terms between 23 and 228? - N. J. A. Sloane, Jul 24 2017
If k > 562 and k! + 1 is divisible by p^2 where p is prime, then either k > 10000 or p > 2038074743 (the hundred millionth prime). - Jason Zimba, Oct 21 2021